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November 17th, 2015, 10:35 PM   #1
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Solve the equation completely.

Solve the equation completely.
$$4^{(2x+3/2)}-65(4^x)+8=0$$
How do I tackle this?

Last edited by Chikis; November 17th, 2015 at 10:55 PM.
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November 17th, 2015, 11:42 PM   #2
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4^(2x + 3/2) - 65(4^x) + 8 = 0

((4^2x) * 4^(3/2)) - 65(4^x) + 8 = 0

[[(4^x)^2] * 8] - 65(4^x) + 8 = 0

let 4^x = a...

Then the expression becomes...

8a^2 -65a + 8 = 0
8a^2 - 64a - a + 8 = 0
8a(a - 8 ) -1(a - 8 ) = 0
(8a - 1)(a - 8 ) = 0
so, a = 1/8 or a = 8

so, a = 4^x = 1/8 or 4^x = 8
4^x = 2^2x = 2^-3 or 2^2x = 2^3

so...

2x = -3 or 2x = 3
x = -3/2 or x = 3/2
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November 18th, 2015, 03:32 AM   #3
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Quote:
Originally Posted by Chikis View Post
How do I tackle this?
Factor out the $4^\frac32 = 8$ in the first term to leave a quadratic in $4^x$.
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November 18th, 2015, 04:10 AM   #4
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Quote:
Originally Posted by Chikis View Post
Solve the equation completely.
Will the real solutions suffice?
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November 18th, 2015, 07:48 AM   #5
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Quote:
Originally Posted by shunya View Post
4^(2x + 3/2) - 65(4^x) + 8 = 0

((4^2x) * 4^(3/2)) - 65(4^x) + 8 = 0 $\displaystyle \ \ \ \ $ * You need "4^(2x)" instead.

[[(4^x)^2] * 8] - 65(4^x) + 8 = 0

let 4^x = a...

Then the expression becomes...

8a^2 -65a + 8 = 0
8a^2 - 64a - a + 8 = 0
8a(a - 8 ) -1(a - 8 ) = 0
(8a - 1)(a - 8 ) = 0
so, a = 1/8 or a = 8

so, a = 4^x = 1/8 or 4^x = 8
4^x = 2^2x = 2^-3 or 2^2x = 2^3 $\displaystyle \ \ \ \ $** You need "2^(2x)" in two places instead.

so...

2x = -3 or 2x = 3
x = -3/2 or x = 3/2


* $\displaystyle \ \ \ \ $4^2x = (4^2)x $\displaystyle \ \ \ \ $You don't mean that.

** $\displaystyle \ \ $2^2x = (2^2)x $\displaystyle \ \ \ \ $You don't mean that.
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November 30th, 2015, 06:55 PM   #6
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